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Suppose one sets up a double-slit experiment, but with only ONE detector behind one of the slits (the left one, let's say). Now suppose the dots that appear on the screen are ERASED (by the experimenter) WHEN the detector noticed a particle going through the left slit. This means the pattern that is formed on the screen after a while consists of dots from UNDETECTED particles only.

Now I can argue in two ways:

1) One can conclude that the particles went through the right slit (since the detector did not notice a particle in the left slit), so there will be no interference pattern.

Or

2) Apparently there wasn't a significant interaction between detector and particles (since the detector did not see them going through the left slit), so there will be an interference pattern.

My questions are:

  • Will there be an interference pattern or not?
  • How does this relate to wavefunction collapse and QM measurements?

EDIT:

My original post above has been marked as unclear. I don't know what was unclear, but I will try to clarify the experiment I had in mind and rephrase the second question which was kind of vague.

Experiment description:

In a double-slit experiment, one can send particles (e.g., electrons) one at a time towards some material with two slits in it. Behind that is a screen on which dots appear when a particle hits it. Suppose two detectors are placed (e.g., strong light sources), one for each slit. When the detectors are turned off, an interference pattern will form eventually on the screen after a large number of particles has been sent. When the detectors are turned on, they interact with the particle wavefuntions and the slit through which a particle goes is identified. Then, after sending many particles, the pattern on the screen does not show interference. The presence of the detectors changes the wavefunctions of the particles. Now I want to understand better how the interactions work. I thought of an experiment with just ONE active detector, that observes, without loss of generality, the left slit only. It may detect a particle or it may not. For each dot that appears on the screen, the experimenter checks whether or not the detector noticed a particle. He keeps track of all the dots that appear when the detector did NOT see a particle go through the left slit (e.g., by marking them in some way, or writing down their screen coordinates, for example).

Now for these marked dots, the detector did not see particles going through the left slit, hence we can conclude that the particles went through the right slit. Since we know through which slit the particles went (the right one), the answer to my original first question is: No, the marked dots will not show an interference pattern.

Now about my second question:

I thought I understood that interactions can collapse wave functions. But what I don't understand in this experiment with one detector, what the interaction is. For the marked dots, the detector did not see particles going through the left slit, so does this mean there was no interaction?

But still the waveform collapsed, since there is no interference pattern. So what was the interaction that caused that? Why did it collapse?

Btw, today I found a similar question here: http://physicshelpforum.com/quantum-physics/12955-double-slits-experiment-one-two-detectors.html

But the question isn't answered.

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closed as unclear what you're asking by Norbert Schuch, ZeroTheHero, Buzz, Kyle Kanos, Jon Custer Jan 7 at 14:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Welcome To Physics SE. Your question is unclear. According to you, the detector detects undetected particles. $\endgroup$ – harshit54 Jan 6 at 11:08
  • $\begingroup$ there is this recent question with more complicated what ifs bu my answer applies here also. physics.stackexchange.com/q/452359 $\endgroup$ – anna v Jan 6 at 11:54
  • $\begingroup$ Thanks Anna. I also expected no interference pattern, but what about my 2nd question? The wavefunction apparently can collapse without an explicit detection. Isn't that strange? $\endgroup$ – anoniem Jan 6 at 16:53
  • $\begingroup$ Thanks, harshit54. By "detector" I don't mean the screen. Is that what was unclear? $\endgroup$ – anoniem Jan 6 at 18:40
  • $\begingroup$ I edited my post to make it more clear. How long will it stay on hold? $\endgroup$ – anoniem Jan 8 at 13:01
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Assume that the detector is 100% accurate, no false positives or negatives. Then if it does not detect a photon you can be sure that it went through the other slit. Thus the detector in fact detects through which slit the photon goes. You economised one detector but this setup still destroys any interference.

See also my answer here for further explanation.

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  • $\begingroup$ I'm not sure it's good practice to assume it went through the other slit, as it could in principle have been absorbed by the blocking material that has a slit in it. $\endgroup$ – StephenG Jan 6 at 15:52
  • $\begingroup$ Thanks my2cts. I also expected no interference pattern, but what about my 2nd question? The wavefunction apparently can collapse without an explicit detection. Isn't that strange? $\endgroup$ – anoniem Jan 6 at 16:54
  • $\begingroup$ Stephen G, if it is absorbed by the blocking material there won't appear a dot on the screen, so there is no problem. $\endgroup$ – anoniem Jan 6 at 16:55
  • $\begingroup$ I explain the bit about the "collapse" in the link I added. $\endgroup$ – my2cts Jan 6 at 17:26
  • $\begingroup$ I think I get it. $d_L$ and $d_R$ are random binary numbers (bits) and $d_L=1-d_R$, right? $\endgroup$ – anoniem Jan 6 at 17:58

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